Maths challenge No. 3

[Wangel]

50 50

 

[cxim]

still waiting for Xede.

Hey man, one of your math profs would make a good judge !

& to make it more fair you can have even money. $100. to $100.

 

[br0wn]

hey cxim, I'll take that bet
But I don't want to take your money.

I'll even take the bet if it is $1 mill to $100 for you.
(means that if you win you get $1 mill, but tell you
the truth I don't have $1 mill, but I know for sure that
I won't lose).

How about you check with your professors or
you math teachers, then come back and admit
that you are wrong ?

 

[cxim]

br0wn,

Ok but your math prof is the judge. Same odds I offered before $100. bucks even.

I haven't been in school for a long time.

 

[br0wn]

cxim, as i said before I DON'T to take your money because
it is not fair, since I know the answer already.

Maybe I should tell you that I helped Math Department at my univ
to teach "Game Theory", "Probability" and even some more advanced
courses before.

 

[Gustavus]

I've never understood why this puzzle causes so much confusion to so many people. Here is a simple analysis. There are two possiblities after your initial choice; either there is a goat behind the door you chose or there is a car. The probability it is a goat is 2/3. The probaility it is a car is 1/3. If a goat is behind the door you chose, there is only one door Monty can open to show you a goat and the car is behind the other door. In either of these two cases, switching will win the car with certainty. If you had already chosen the door with a car behind it, Monty can open either door to reveal a goat, but there is a goat behind the other door so switching looses with certainty. The probabilty you will win the car if you switch is therefore:

2/3x1 + 1/3x0 or 2/3.

 

[cxim]

br0wn

your history does not impress me. I will put up $200 to your $100.

We can work out the details of how. You can run the test. All I require is an honest observer.

 

[br0wn]

You are so stubborn, aren't you ?

I don't want your money

How about you tell me what OS you are running, and I write
a simulation program for you and you can verify for yourself ?

 

[cxim]

no simulation.....

you do the test !

tell you what, where do you live ? we will do the test together !

Boston was looking good but Xede kind of disappeared.

here is the deal. I will come up for a few days, if I am right, you pay for the trip, after the test.

If you are right, i pay my own way & you get a trip to Florida ?

 

[Mister T]

People, people, people......

brOwn is correct. Cxim, I would not bet if I were you.
I have done math problems like this in the past and brOwn is correct.
The reasoning is in this thread so I will not go over it again.

By the way, I was nationally ranked in mathematics a few years ago
and currently a senior at MIT so I am positively sure that brOwn is correct.

Yo brOwn, where is my 20% cut?

 

[tom3]

This was probably mentioned earlier in the thread, but it is the most obvious way to see this. The underlined letter is your current selection, G = goat, C = car and X = elminated goat.

Here are ALL the possible cases:

GGC
GGC
GGC

GCG
GCG
GCG

CGG
CGG
CGG

The 3 blocks are actually reiterations, since the only difference is the arrangement of the prizes. They are shown here for the sake of completeness.


If you decide to stay with the door you picked initially:

GGC -> XGC or GXC --you win the car
GGC -> XGC -----------------you win a goat
GGC -> GXC -----------------you win a goat

GCG -> XCG -----------------you win a goat
GCG -> XCG or GCX --you win the car
GCG -> GCX -----------------you win a goat

CGG -> CXG -----------------you win a goat
CGG -> CGX -----------------you win a goat
CGG -> CXG or CGX --you win the car

You win the car if you initially picked the car, and you get a goat if you initially picked a goat. It doesn't matter at all whether you are shown a goat or not. Therefore the probability of you winning a car is a straight forward 1/3, and winning a goat 2/3.


If you decide to switch after you are shown a goat:

GGC -> XGC or GXC -> XGC or GXC --you win a goat
GGC -> XGC -> XGC -----------------you win the car
GGC -> GXC -> GXC -----------------you win the car

GCG -> XCG -> XCG -----------------you win the car
GCG -> XCG or GCX -> XCG or GCX --you win a goat
GCG -> GCX -> GCX -----------------you win the car

CGG -> CXG -> CXG -----------------you win the car
CGG -> CGX -> CGX -----------------you win the car
CGG -> CXG or CGX -> CXG -----------------you win a goat

You win the car if you initially picked a goat, and win a goat if you initially picked the car. Since we all agree that the chances of you picking the car initially is 1/3, and the chances of you picking a goat is 2/3, we can conclude that you will win a goat 1/3 of the time, and win the car 2/3 of the time.


I really hope this would be helpful to some of you so that you can see the truth.

 

[cxim]

Ah... tom3....it was a good bluff for a while

 

[br0wn]

For those you NON-BELIEVER of SWITCHING,

Try this program, it should run on any Windows machine :

1. This is the executable, rename it to monty.exe (since geocities
won't allow me to upload .exe file)
"Executable file, but rename it to monty.exe"

2. This is the source code, compile this for yourself if you don't
trust my binary code
"Source code, you need a C compiler to compile this"

edit :
The program allows YOU to play the Monty's game for as many time as you like
and keep track of your winning.
I suggest to play more than 10 times, so you can get quite accurate probability.
Try always switch strategy, and always NOT switch strategy, and compare the two, which
is better.

edit no 2 :
Mister T, sure I can give you 20% of the winning, but I don't want to bet (coz I'd feel
like taking candy from a little girl ).

 

[callspread2]

Sorry I disappeared for a while - I lost my password and can't login as callspread for a while.

I won't take your bet - howabout instead you give me $1000 and I give you $833 in return.

The probability of the car being behind one of the doors is always 1:1, regardless of how many doors there are.
Given three doors, door A, door B and door C, I randomly choose door B. The probability of finding the car = 1, therefore A+B+C = 1.
The host then eliminates a door, say door C, leaving doors A and B. Since the probability of finding the car is still 1 for both the doors together, A+B = 1. Now, since you say that I will improve my chances by switching, door A must have a greater probability than door B. A>B. You claim, correctly, that the chance for door A (not my choice) is 1/2, therefore B<1/2. Specifically, you claim that B=1/3. Since A+B=1, and A+B=0.83, therefore 0.83=1.

 

[Duckers]

Let it die guys.

1st event
GOAT
GOAT
CAR

Chance to get the car = 1/3

2nd event
GOAT
CAR

Monty asks to decide whether or not we want to switch or stick with the door we picked. At this point the chance is simply 1/2.

Let's get over it and move on to Math Challenge #4!

 

[br0wn]

Duckers, why don't you try my program ?

here :
&quot;Executable, but rename it to monty.exe&quot;


Ok, this is my last post in this thread, I'm tired of arguing
with those who FOR A SEC doesn't even want to look/read at what
the others say.

 

[Duckers]

I did read what other people had to say.

Both arguments are valid, they just depend on how you want to analyze the question.

 

[allan120]

callspread and Duckers, your arguments are only valid if Monty randomly switches the items behind the doors after he opens one door. You assume that the events are independent, when they are actually related.

In callspread's equations, his second equation A+B=1 is missing C. The equation after a door is opened is A+B+C=1, only we know that the .33 chance that C is holding the car is no longer applicable. The proper equation after a door is opened is A+B=0.66, where A = 0.33 and B=0.33.

Note that although A=B=0.33, the odds after a door is opened are not 50/50 as long as the items are not switched randomly behind the closed doors.

Oh well, I have to pick up my sister so I don't have time to finish my argument, but your equation is flawed. We could leave it as Duckers suggested, but then we'd be leaving it as incorrect. His diagram for the second pick &quot;Goat vs. Car&quot; is oversimplified and disregarding the third door, much like callspread's equation.

 

[Duckers]

callspread and Duckers, your arguments are only valid if Monty randomly switches the items behind the doors after he opens one door. You assume that the events are independent, when they are actually related.

Monty does not need to switch the items behind the 2 doors left.

The car could be in either door.

 

[kranky]

duckers, I don't think you believe that &quot;both arguments are valid&quot;. We want you to see the right answer, it's not being argumentative. I see people disagreeing, but not refuting the logic of the correct explanations.

For those who believe the odds are 50/50:
Pick a door and NEVER switch. Since you say the odds are 50-50, then it makes no difference whether you switch or not. That is saying you have a 50-50 chance of picking the right door the first time. But we all agreed it is a 1/3 chance.

I am looking forward to Argo's post when he simulates the outcomes later.

And just the fact that Gustavus says switching is better should be the clincher, even if the logic is complicated.

 

[Duckers]

His diagram for the second pick &quot;Goat vs. Car&quot; is oversimplified and disregarding the third door, much like callspread's equation.


Actually, The third door must be disregarded.

The contestant then chooses a door. Monty then opens a different door, revealing a goat.

There is a goat in that door. YOU KNOW there is a goat in that door because Monty is showing you the Goat.
Therefore the third door is elimated.

 

[Duckers]

That is saying you have a 50-50 chance of picking the right door the first time.

Theorically yes, Monty is going to eliminate one of the goats anyway.

 

[Mister T]

CXIM,

If I can get a MATH professor @ MIT (Massachusetts Institute of Technology)
to sign a proof that I write as being correct, that confirms a switching strategy will
win 2/3 of the time, will you bet me money?

I would be able to provide all contact info for myself and the professor.
Does this sound like a reasonable bet?

I am ready to bet any amount of money under $10,000 USD.

Let me know,
Tony

 

[Souka]

Good god some of ya are idiots....I would lock this thread if I could.

My Two Grumpy Pesos...

 

[allan120]

<< That is saying you have a 50-50 chance of picking the right door the first time. Theorically yes, Monty is going to eliminate one of the goats anyway. >>

Okay Duckers, this statement, that you initially have a &quot;theoretical&quot; 50-50 chance because Monty is eliminating all but one goat, is logically the same as saying you have a 50-50 chance picking the car out of 1 million doors, if Monty then eliminates all but one goat. If you can convince anyone that an initial selection out of a million doors is a 50-50 chance, please provide a mathematical proof.

I'm looking forward to Argo's post as well.